Ch 17 Thermochemistry

What is the difference between heat and temperature? HEAT is energy that transfers from one object/substance to another TEMPERATURE is a measure of the amount of energy an object/substance has

A. Energy Transformations Energy that is stored in the chemical bonds of a substance is called CHEMICAL POTENTIAL ENERGY Heat ALWAYS flows from hot to cold

B. Endothermic and Exothermic Processes System=the reaction Surroundings=everything around the reaction Surroundings Law of Conservation of Energy- energy can be neither created nor destroyed System

B. Endothermic and Exothermic Processes Endothermic Process- heat is absorbed from the surroundings Endo = Into Endothermic processes are represented by a positive “q” HEAT

B. Endothermic and Exothermic Processes Exothermic process- heat is released into the surroundings Exo = Exit Exothermic processes are represented by a negative “q” HEAT

C. Measuring Heat Flow Two Common Units 1J = 0.2390 cal Joule calorie 1J = 0.2390 cal 4.184 J = 1 cal 1Calorie = 1 kilocal = 1000 cal

D. Heat Capacity and Specific Heat Heat Capacity depends on: The mass of the object The chemical composition of the object “the amount of heat needed to increase the temperature of an object by 1 oC Specific heat capacity- amount of heat needed to raise the temperature of 1g by 1 oC

D. Heat Capacity and Specific Heat C = q / (m X ΔT) C =Specific Heat q = heat (joules or calories) m = mass (grams) ΔT = change in temperature The change in temperature can be measure in Kelvin or degrees Celsius

C= 849J/ (95.4g)(23.0 oC) = 0.387 J/(g· oC) The temperature of a 95.4g piece of copper increases from 25.0 oC to 48.0 oC when it absorbs 849 J of heat. What is the specific heat of copper? Known: m= 95.4g q=849 J ΔT= 48.0-25.0=23.0 oC Work: C= 849J/ (95.4g)(23.0 oC) = 0.387 J/(g· oC)

E. Calorimetry Measures the heat flow into or out of a system Heat released by the system is equal to heat absorbed by the surroundings ENTHALPY: (H) the heat constant of a system at constant pressure

E. Calorimetry The terms heat and enthalpy change are interchangeable q = ΔH qsys = ΔH = -m x C x ΔT Negative enthalpy = exothermic Positive enthalpy = endothermic

When 25. 0mL of water containing HCl at 25. 0 oC is added to 25 When 25.0mL of water containing HCl at 25.0 oC is added to 25.0mL of water containing NaOH at 25.0 oC in a calorimeter a rxn occurs. Calculate the enthalpy change (in kJ) during the rxn if the highest temperature observed was 32.0 oC. Assume all densities =1g/mL KNOWN: Cwater=4.18J/g oC V=25.0mL+25.0mL ΔT=7 oC Density= 1g/mL ΔH = ?

m= (50mL) x (1. 00g/mL) = 50g ΔT= TF – Ti = 32. 0 – 25. 0 =7 m= (50mL) x (1.00g/mL) = 50g ΔT= TF – Ti = 32.0 – 25.0 =7.0 ΔH= -mCΔT= -(50.0g)(4.18J/goC)(7.0oC) ΔH= -1463 J = -1500J = -1.5kJ

F. Thermochemical Equations In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product Endothermic (positive ΔH) 2NaHCO3 + 129kJ Na2CO3 + H2O + CO2 Exothermic (negative ΔH) CaO + H2O Ca(OH)2 + 65.2kJ

G. Heat of Combustion The heat of reaction for the complete burning of one mole of a substance Written the same was as change in enthalpy

Write the thermochemical equation for the oxidation of Iron (III) if its ΔH= -1652 kJ Fe(s) + O2(g)→ Fe2O3(s) + 1652 How much heat is evolved when 10.00g of Iron is reacted with excess oxygen? 2 4 3 10.00g Fe 1 mol = 0.1791 mol Fe 55.85g Fe 0.1791 mol Fe 1652 kJ =73.97 kJ of heat 4 mol Fe